Brain Teasers at Xmas
Posted by: Don Atkinson on 23 December 2010
Brain Teasers at Xmas
Xmas time - and many of us will be off on holiday or meeting up with friends and family. For those with a bit of time on their hands, I thought I would post a dozen (12 days of Xmas ?) brain teasers to help keep your minds in gear.
All are taken from the Brain Teaser thread from a few years back and credit is given to the original poster where this is other than me. Ken c still posts here, but BAM and Matthew T….??
None of these teasers are “trick” questions, but you will need to read them carefully in order to understand some of them. Discuss at leisure......
I know it’s not as exciting as today’s hi-tech electronic games, but I hope it helps to keep some of you amused over Xmas, and as Ken c would say – “enjoy”
Cheers
Don
Posted on: 18 January 2011 by Don Atkinson
Ian, Fred
I will get back properly once i've sorted out this new Forum format.
Meanwhile,
Ian - you've got the right answer
Fred - its going to take a while to draft a helpful response
cheers
Don
Posted on: 20 January 2011 by Don Atkinson
Ian,
The answer to Q7 is (definitely) 36pi – well done!
Did you rely on the “assumption” that if “R” is not given, it obviously isn’t necessary?
Or did you “prove” this assumption first?
Cheers
Don
Posted on: 20 January 2011 by Don Atkinson
Fred,
“Now please don’t tell me its wrong”……I won’t, you’ve made a good start, but……….
Let me gently point out, the challenge requires that you identify the specific ball that is the odd one out AND whether that ball is heavier or lighter than the other balls.
Your first two weighings have made certain (unjustified?) assumptions and they don’t give you enough information to solve the problem.
(1) aaaa = bbbb
(2) aaaa > cxcc
These two weighings tell you that:-
(1) the odd ball is not aaaa or bbbb. (It must therefore be one of the other 4 balls)
(2) the odd ball is lighter than a standard ball
However, you don’t know which of the 4 balls on the RHS in weighing (2) is the odd ball. They all LOOK the same.
Also worth noting, if (and this is a possibility) the odd ball was one of the 8 balls in your first weighing (ie it was aaaa or bbbb) the scales wouldn’t balance! However, you would then at least know the odd ball was either one of the aaaa balls, or one of the bbbb balls and definitely NOT one of the cccc balls.
Nonetheless, as I said above, you have made a very good start. My solution starts with putting 4 balls on the left scale-pan and 4 balls on the right scale-pan and observing the result. This is the first weighing. You only need two more to identify the odd ball AND decide whether it is light or heavy.
Cheers
Don
Posted on: 20 January 2011 by Adam Meredith
Shirley - I gave you this before.
Posted on: 20 January 2011 by Don Atkinson
Adam you fiend, you did!!
In fact all the brain teasers in this thread have been answered before on the forum, as I think I made clear at the begining. But to keep some of the newer members out of trouble over Xmas, I thought they might like to try a few of them themselves.
Fred has obviously had a shot at a couple, but seemed to be in need of a bit of help with this one.
Problem now solved! - Fred?
Cheers
Shirley
Posted on: 21 January 2011 by Fred Mulder
Don/Adam thanks, I'll read and respond later.
Cheers, Fred
Posted on: 21 January 2011 by Fred Mulder
Don/Adam thanks, I'll read and respond later.
Cheers, Fred
Posted on: 24 January 2011 by Ian G.
Don,
A bit of both, used the lack of a given radius to hint it might not be necessary, but then worked out the volume of the drilled out section (cylindrical core + two identical end cap sections) from the 6inch length constraint. This showed the leftover part is indeed independent of the radius which is neat.
Not sure how one could do it without knowing some calculus though, so is there a cute geometry based shortcut I'm missing?
cheers
Ian
Posted on: 25 January 2011 by Fred Mulder
D'oh..!! had a look at Adams attach.. And I thought this one was quite simple, but you're cheating with maths again are you Don? Well, you asked for the MINIMUM, and I gave a possible correct outcome. For me it's sorted
Posted on: 25 January 2011 by Don Atkinson
"Not sure how one could do it without knowing some calculus though, so is there a cute geometry based shortcut I'm missing?"
Ian,
When I first saw this particular teaser, I also assumed the lack of the radius indicated the answer was independent and quickly got 36pi. Even if you started off with an 8,000 mile dia sphere (eg the world)
Eventually, i put the old brain into gear and did the calculus.
i don't know of any cute geometric solutions - sorry!
Cheers
Don
Posted on: 25 January 2011 by Don Atkinson
"D'oh..!! had a look at Adams attach.. And I thought this one was quite simple, but you're cheating with maths again are you Don?"
Fred,
No maths in my solution, just logic.
I'll see if I can attach my solution (i'm still struggling with this new forum - so it will be a separate post!!!). I think (hope) you might find it easier to grasp than Adam's. There is nothing wrong with Adam's solution but I think my solution is a bit more intuitive and lets you visualise how three weighings are needed to guarantee success. It also shows how you might be lucky and get the result in only two weighings.
Give me a day or two - i'm just a little busy at work with night flying at present.
Cheers
Don
Posted on: 26 January 2011 by Don Atkinson
Fred
My solution for finding the odd ball (hopefully)
Note, you need 3 weighings to guarantee to find the odd ball.
Cheers
Don
Posted on: 26 January 2011 by Don Atkinson
Well.................
The attachment in the previous post isn't brilliant. I can read it, but then I wrote it.............
Cheers
Don
Posted on: 26 January 2011 by Ian G.
ok an inelegant formula for #8. The ladder should be placed a distance d from the crate where d satisfies:
d = cot ( acos( (1+d)/L ) )
Ugly transcedental equation that seems to work. I can also produce an even uglier quartic equation for d.
I await the elegant solution ......
Ian
..and shortest ladder that can be used is 2*sqrt(2) meters long!
Posted on: 27 January 2011 by Ian G.
and for number 9 since no-one seems to have written it out.
n(n-1)(n-2)(n-3) = n(n-3)(n-2)(n-1)
= (n^2 - 3n)(n^2-3n+2)
= (m-1)(m+1) where m= n^2-3n+1
=m^2-1
et voila
Since we're worshipping the loveliness of exp(i pi)=-1 have any of you ever worked out the value of i^i (in words i to the power i) The result might surprise you-it did me.
Ian
Posted on: 27 January 2011 by Don Atkinson
"d = cot ( acos( (1+d)/L ) )"
Ian,
you seem to have "d" on both sides of the equation........??
Cheers
Don
Posted on: 27 January 2011 by Ian G.
yes, that is what a transcedental equation is e.g. x=sin(x), valid but ugly. I guess you are saying there is a nicer solution so I'll keep trying.......
Ian
Posted on: 28 January 2011 by Don Atkinson
"I guess you are saying there is a nicer solution so I'll keep trying......."
Ian,
I had forgotten the term "trancendental equation" - thanks for the reminder!
It is possible to derive an algebraic equation that provides a solution for "d" purely in terms of "L" without the need for iteration.
The basic equation is pig ugly. It can be made somewhat elegant-looking simply by making a substitution eg Let "y" = "the pig ugly bit" of the RHS of the equation!. That is the form of my solution. You can calculate "d" with a basic calculator, log tables etc. The pig-ugly bits are dead easy to calculate.
Ken c managed to derive a (slightly IMHO!!!!) more elegant form of equation. You need a more sophisticated set of log tables to calculate "d"
Good luck - it might be worth sharing your progress here to see if others can help out.........
Cheers
Don
Posted on: 28 January 2011 by Ian G.
hi,
I know there is a 'pig ugly' solution to quartic equations so I guess that might be what you are looking for. I'll type it out later. Whether that is more or less elegant than a short transcedental equation I leave as a debating point for the pub one day.
Posted on: 28 January 2011 by Don Atkinson
Ian
iteration is obviously valid, and well done for finding your equation. nobody came up with that format last time round.
Even L^2=d^2 + 2d + 2 + 2/d + 1/(d^2) could be used. Its a fairly elegant looking equation (well, IMHO). The convergence might not be rapid, but it certainly works.
Your transedental equation probably converges far more rapidly (i haven't tried it!).
Some quartic equations have easier solutions than others. But those that do have solutions (and the Ladder does) at least give an answer (or both solutions in the case of the Ladder) without the need for iteration.
As I said above, Ken c did find a slightly more elegant form of the equation. Bam, who set the puzzel, left the forum without indicating whether Ken's solution was elegant enough!! Bam wasn't impressed with my offering!! plenty of scope for others to succeed..........
Cheers
Don
Posted on: 29 January 2011 by Don Atkinson
"Even L^2=d^2 + 2d + 2 + 2/d + 1/(d^2) could be used."
This requires inelgant "trial & error" rather than the more sophisticated "iteration" proposed by Ian.
Either way, you can plot the curves and use them to find solutions. Clumsey, inelegant but.......it works.
So we are looking for something that doesn't require trial & error or iteration. 
Cheers
Don Posted on: 30 January 2011 by Don Atkinson
Re-reading my last post, the wording seems abrupt and gives the appearance of being critical of others. I had no idea when I wrote it, that it might be read in such a light. I was simply trying to indicate that Ian's iterative solution was a lot better than the "trial & error" option that I had posted before.
My sincere apologies if I have offended anybody.
Cheers
Don
Posted on: 31 January 2011 by Ian G.
Certainly haven't offended me Don
You can easily go from my solution to the polynomial one using the trig identity:
sin(acos(x)) = sqrt(1 - x^2)
and a bit of re-arrangement so they are both exactly equivalent.
The solution to the quartic is straightforward but longwinded so I won't try and type it out!
cheers
Ian
Posted on: 01 February 2011 by Don Atkinson
"The solution to the quartic is straightforward but longwinded so I won't try and type it out!"
My version is probably the same as yours. Let me try......
2d = y - sqrt ( [y^2] - 4) where y = "a very simple function of L"
Note: I have used 2d rather than d, simply to avoid having to say "all the RHS/2"
Ken c publised a solution.....
d = e^-Acosh(z) where z = "another very simple function of L"
Ring any bells??
Cheers
Don
PS - i'd better find and check my original calcs, i've just done the above quickly and they sort of look right.
Posted on: 01 February 2011 by Ian G.
Hi Don,
Yep I have the same as your solution. The substitution y=d+1/d in the quartic reduces it to a wee quadratic for y. Solving this for y as a function of L, allows one to next solve for d from the substitution equation y=d+1/d which is also a quadratic. So y=-1 + sqrt(1+L^2) in your solution.
Ain't maths grand?
Have you figured out the value of i^i as I posed earlier?
Ian
